A circle \({x^2} + {y^2} + dx + ey + c = 0\) and a straight line \(lx + my + n = 0\) intersect. Find the general equation of a circle which passes through the points of intersection, justifying your answer.

METHOD 1

\({x^2} + {y^2} + dx + ey + c + \lambda (lx + my + n) = 0\)     M1 A1

ie\(\;\;\;\)\({x^2} + {y^2} + x(d + \lambda l) + y(e + \lambda m) + c + \lambda n = 0\)     A1

since \({x^2}\) and \({y^2}\) have the same coefficients and there is no \(xy\) term, this is a circle     R1

we know the pair of points fit the equation.     R1

hence this is the required equation.

METHOD 2

Let the general equation be

\({x^2} + {y^2} + ax + by + q = 0\)     M1

The intersection with the given circle satisfies

\((a - d)x + (b - e)y + (q - c) = 0\)     M1A1

This must be the same line as \(lx + my + n = 0\)     R1

Therefore

\(a - d = \lambda l\;\;\;{\text{giving}}\;\;\;a = d + \lambda l\)

\(b - e = \lambda m\;\;\;{\text{giving}}\;\;\;b = e + \lambda m\)     A1

\(q - c = \lambda n\;\;\;{\text{giving}}\;\;\;q = c + \lambda n\)

leading to the required general equation

Note: Award M1 to candidates who only attempt to find the points of intersection of the line and circle

This was the worst answered question on the paper and indeed no complete solution was seen with no candidate having the required insight to write down the solution. Most candidates who attempted the question tried to find the points of intersection of the line and circle which led nowhere.